Manipal Entrance Test 2014 Math Paper

Given system of equation is
x+y+z=6
x+2y+3z=10
x+2y+λ=µ
If given system of equation has no solution, then D = 0 and atleast one of the determinants
D1,D2,D3≠0
Here‌‌‌D=0
⇒ |

1	1	1
1	2	3
1	2	λ

|=0
⇒ 1(2λ−6)−1(λ−3)+1(2−2)=0
⇒ 2λ−6−λ+3+0=0
⇒ λ−3=0
⇒ λ=3
Also, D1=|

6	1	1
10	2	3
µ	2	3

|≠0
⇒ 6(6−6)−1(30−3µ) +1(20−2µ)≠0
⇒ 0−30+3µ+20−2µ≠0
⇒ µ−10≠0
⇒ µ≠10